Let $f(n) = \sum_{k=1}^{\lfloor \frac{n}{3} \rfloor} \binom{n}{3k} 8^k$ and $\omega \neq 1$ be a third root of the unity. Using the fact that for every integer $k \geq 0$:
$$1 + \omega^k + \omega^{2k} = \begin{cases} 3, & \text{if } 3 \mid k \\ 0, & \text{otherwise,} \end{cases}$$
we get that:
$$\begin{aligned} f(n) + 1 &= \sum_{k=0}^{\lfloor \frac{n}{3} \rfloor} \binom{n}{3k} 2^k = \frac{1}{3} \sum_{k=0}^{n} (1 + \omega^k + \omega^{2k}) \binom{n}{k} 2^k \\ &= \frac{1}{3} 3^n + \frac{1}{3} (1 + 2\omega)^n + \frac{1}{3} (1 + 2\omega^2)^n. \end{aligned}$$
Now note that $3, 1+2\omega$ and $1+2\omega^2$ are the roots of the polynomial:
$$P(x) = (x-3)(x-1-2\omega)(x-1-2\omega^2) = (x-1)^3 - 8 = x^3 - 3x^2 + 3x - 9$$
which, in turn, is the characteristic polynomial of the recursive sequence $(a_i)_{i \geq 0}$:
$$a_{i+3} = 3a_{i+2} - 3a_{i+1} + 9a_i \text{ for } i \geq 0.$$
Thus, if we set $a_i = f(i) + 1 = 1$ for $0 \le i \le 2$, then $f(n) + 1 = a_n$ for every $n \ge 0$. Let $b_i = a_i \pmod{10^{2023}}$. Since $\text{gcd}(3, 10^{2023}) = 1$, any three consecutive terms of the sequence $(b_i)_{i \ge 0}$ uniquely determine the previous as well as the next term of this sequence. Together with the fact that there are only finitely many residues modulo $10^{2023}$, we conclude that the sequence $(b_i)_{i \ge 0}$ is periodic with some period $d > 3$ (since $b_3 = a_3 = 9$). Therefore:
$$9(f(d-1)+1) = 9a_{d-1} = a_{d+2}-3a_{d+1}+3a_d \equiv a_2-3a_1+3a_0 \pmod{10^{2023}} = 1 \pmod{10^{2023}}.$$
Finally, since $9 \mid 8.10^{2023} + 1$, we conclude that $9^{\frac{8.10^{2023}+1}{9}} \equiv 1 \pmod{10^{2023}}$ and consequently:
$$f(d-1) + 1 = a_{d-1} \equiv \frac{8.10^{2023} + 1}{9} = \underbrace{88 \dots 89}_{2022} \pmod{10^{2023}}$$

and thus $f(d-1) \equiv \underbrace{88 \dots 8}_{2022} (\text{mod } 10^{2023})$. Therefore $n = d-1$ has the desired property. $\square$